Numbers of form 31 x 16^n are sum of 16 4th Powers
Theorem
Let $m \in \Z$ be an integer of the form $31 \times 16^n$ for $n \in \Z_{\ge 0}$.
Then in order to express $m$ as the sum of fourth powers, you need $16$ of them.
Proof
We have:
- $31 \times 16^n = \paren {2^{n + 1} }^4 + 15 \times \paren {2^n}^4$
so every integer of the form $31 \times 16^n$ for $n \in \Z_{\ge 0}$ can be expressed as the sum of 16 fourth powers.
Now we show that we cannot use less than $16$ fourth powers.
Observe that for an even number $2 k$:
\(\ds \paren {2 k}^4\) | \(=\) | \(\ds 16 k^4\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds 0\) | \(\ds \pmod {16}\) |
For an odd number $2 k + 1$:
\(\ds \paren {2 k + 1}^4\) | \(=\) | \(\ds 16 k^4 + 32 k^3 + 24 k^2 + 8 k + 1\) | Binomial Theorem | |||||||||||
\(\ds \) | \(\equiv\) | \(\ds 8 k \paren {k + 1} + 1\) | \(\ds \pmod {16}\) | |||||||||||
\(\ds \) | \(\equiv\) | \(\ds 1\) | \(\ds \pmod {16}\) | since $k \paren {k + 1}$ is even |
It is obvious that for $n = 0$, $31$ requires $16$ fourth powers to express:
- $31 = 2^4 + 15 \times 1^4$
Aiming for a contradiction, suppose for some $n > 0$, $31 \times 16^n$ requires less than $16$ fourth powers to express.
Let $m$ be the smallest of those $n$.
Suppose $x$ of the summands are odd, where $x < 16$.
By the above, we must have:
- $31 \times 16^m \equiv x \pmod {16}$
Since $31 \times 16^m$ is divisible by $16$, $x = 0$.
Hence each summand is even.
Dividing each summand by $2$ gives a representation of $31 \times 16^{m - 1}$ as a sum of less than $16$ fourth powers.
This contradicts the minimality condition on $m$.
Hence the result by Proof by Contradiction.
$\blacksquare$